$\int^{1/4}_{0}\dfrac{8x}{\sqrt{1-4x^2}}\,dx\, = $
Answer: Strategy Let's first find the indefinite integral $\int\dfrac{8x}{\sqrt{1-4x^2}}\,dx\, $. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\dfrac{8x}{\sqrt{1-4x^2}}\,dx\, $, we can use U-substitution. If we let $ {u=1-4x^2}$, then ${du=-8x \, dx}$ and ${ 8x\,dx=-du}$. So we have: $\begin{aligned}\int\dfrac{8x}{\sqrt{1-4x^2}}\,dx\, &=\int\dfrac{1}{\sqrt{{1-4x^2}}}\,\cdot{8x\, dx}\,\\\\\\\\ &=\int\dfrac{1}{\sqrt{ u}}\,\cdot {-1\, du}\,\\\\\\\\ &=-1\int\dfrac{1}{\sqrt{u}}\,du\\\\\\\\ &=-1\cdot 2\sqrt{u}+C\\\\\\\\ &=-2\sqrt{1-4x^2}+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{1/4}_{0}\dfrac{8x}{\sqrt{1-4x^2}}\,dx\, &= -2\sqrt{1-4x^2}\Bigg|^{1/4}_0\\\\\\\\ &=-2\left(\sqrt{\dfrac34}-1\right)\\\\\\\\ &=-2\left(\dfrac{\sqrt{3}}{2}-1\right)\\\\\\\\ &=2-\sqrt{3}\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{1/4}_{0}\dfrac{8x}{\sqrt{1-4x^2}}\,dx\, = 2-\sqrt{3}$